Analytical Geometry - Conics - Ellipse

Find the equation of ellipse whose latus-rectum is $5$ and eccentricity is $\dfrac{2}{3}$.

Latus-rectum of ellipse $= \dfrac{2b^2}{a} = 5$ (given)

$\implies$ $b^2 = \dfrac{5 a}{2}$ $\;\;\; \cdots \; (1)$

Eccentricity $= e = \dfrac{2}{3}$ (given)

For an ellipse, $\;$ $b^2 = a^2 \left(1 - e^2\right)$ $\;\;\; \cdots \; (2)$

Substituting the values of $b^2$ and $e$ in equation $(2)$ gives,

$\dfrac{5a}{2} = a^2 \left[1 - \left(\dfrac{2}{3}\right)^2\right]$

i.e. $\;$ $\dfrac{5}{2} = \dfrac{5a}{9}$

i.e. $\;$ $a = \dfrac{9}{2}$ $\implies$ $a^2 = \dfrac{81}{4}$

Substituting the value of $a$ in equation $(1)$ gives,

$b^2 = 5 \times \dfrac{9}{2} \times \dfrac{1}{2} = \dfrac{45}{4}$

The equation of ellipse is $\;\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$

i.e. $\;$ $\dfrac{x^2}{81 / 4} + \dfrac{y^2}{45 / 4} = 1$

i.e. $\;$ $\dfrac{4 x^2}{81} + \dfrac{4 y^2}{45} = 1$

Analytical Geometry - Conics - Rectangular Hyperbola

Prove that the tangent at any point to the rectangular hyperbola forms with the asymptotes a triangle of constant area.

Let the rectangular hyperbola be: $\;$ $xy = c^2$ $\;\;\; \cdots \; (1)$

The asymptotes of the rectangular hyperbola given by equation $(1)$ are the $X$ and $Y$ coordinate axes.

Any point $P \left(t\right)$ on equation $(1)$ is $\;$ $P \left(t\right) = P \left(ct, \dfrac{c}{t}\right)$

Equation of tangent at point $P \left(t\right)$ is

$x + yt^2 = 2ct$ $\;\;\; \cdots \; (2)$

Putting $y = 0$ in equation $(2)$, we get the coordinates of point $A$ as $A \left(2ct, 0\right)$.

Putting $x = 0$ in equation $(2)$, we get the coordinates of point $B$ as $B \left(0, \dfrac{2c}{t}\right)$

$\begin{aligned} \therefore \; \text{Area of } \triangle AOB & = \dfrac{1}{2} \times OA \times OB \\\\ & = \dfrac{1}{2} \times 2 ct \times \dfrac{2c}{t} \\\\ & = 2c^2 = \text{constant} \end{aligned}$

$\therefore$ $\;$ The tangent at any point to the rectangular hyperbola forms with the asymptotes a triangle of constant area.

Hence proved.

Analytical Geometry - Conics - Rectangular Hyperbola

Find the equations of the asymptotes for the rectangular hyperbola $6x^2 + 5xy - 6y^2 + 12x + 5y + 3 = 0$

Equation of the rectangular hyperbola is

$6x^2 + 5xy - 6y^2 + 12 x + 5y + 3 = 0$ $\;\;\; \cdots \; (1)$

The combined equation of the asymptotes differs from the hyperbola by a constant.

$\therefore$ $\;$ The combined equation of the asymptotes is

$6x^2 + 5xy - 6y^2 + 12x + 5y + k = 0$

Consider

$\begin{aligned} 6x^2 + 5xy - 6y^2 & = 6x^2 + 9xy - 4xy - 6y^2 \\\\ & = 3x \left(2x + 3y\right) - 2y \left(2x + 3y\right) \\\\ & = \left(3x - 2y\right) \left(2x + 3y\right) \end{aligned}$

$\therefore$ $\;$ The separate equations of asymptotes are

$3x - 2y + \ell = 0$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $2x + 3y + m = 0$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ $\left(3x - 2y + \ell\right) \left(2x + 3y + m\right) = 6x^2 + 5xy -6y^2 + 12x + 5y + k$

Equating the coefficients of $x$ terms we get: $\;$ $3m + 2 \ell = 12$ $\;\;\; \cdots \; (3a)$

Equating the coefficients of $y$ terms we get: $\;$ $-2m + 3 \ell = 5$ $\;\;\; \cdots \; (3b)$

Equating the constant terms we get: $\;$ $\ell \times m = k$ $\;\;\; \cdots \; (3c)$

Solving equations $(3a)$ and $(3b)$ simultaneously we get,

$\ell = 3$ $\;$ and $\;$ $m = 2$

Substituting the values of $\ell$ and $m$ in equations $(2a)$ and $(2b)$ respectively, we get the separate equations of asymptotes as

$3x - 2y + 3 = 0$ $\;$ and $\;$ $2x + 3y + 2 = 0$

Analytical Geometry - Conics - Rectangular Hyperbola

Find the equations of the asymptotes for the rectangular hyperbola $2xy + 3x + 4y + 1 = 0$

Equation of given rectangular hyperbola is

$2xy + 3x + 4y + 1 = 0$

i.e. $\;$ $xy + \dfrac{3}{2}x + 2y + \dfrac{1}{2} = 0$

i.e. $\;$ $xy + \dfrac{3}{2}x + 2y + \dfrac{1}{2} + 3 = 3$

i.e. $\;$ $xy + \dfrac{3}{2}x + 2y + \dfrac{7}{2} = 3$

i.e. $\;$ $y \left(x + 2\right) + \dfrac{3}{2} \left(x + 2\right) = 3$

i.e. $\;$ $\left(x + 2\right) \left(y + \dfrac{3}{2}\right) - 3 = 0$

The equation of a rectangular hyperbola differs from the combined equation of asymptotes by a constant.

$\therefore$ $\;$ The combined equation of asymptotes is

$\left(x + 2\right) \left(y + \dfrac{3}{2}\right) = 0$

$\therefore$ $\;$ The separate equations of asymptotes are

$x+ 2 = 0$ $\;$ and $\;$ $y + \dfrac{3}{2} = 0$

Analytical Geometry - Conics - Rectangular Hyperbola

Find the equation of the rectangular hyperbola which has its center at $\left(2,1\right)$, one of its asymptotes $3x - y - 5 = 0$ and which passes through the point $\left(1, -1\right)$.

Let the equation of the required rectangular hyperbola be

$\left(x - h\right) \left(y - k\right) = c^2$ $\;\;\; \cdots \; (1)$

One of the asymptotes of the required rectangular hyperbola is

$3x - y - 5 = 0$ $\;\;\; \cdots \; (2)$

Slope of the given asymptote is $= m_1 = 3$

The asymptotes of a rectangular hyperbola are perpendicular to each other.

$\therefore$ $\;$ Slope of the second asymptote $= m_2 = \dfrac{-1}{m_1} = \dfrac{-1}{3}$

Let the equation of the second asymptote be

$y = \dfrac{-1}{3}x + p$ $\;$ where $p$ is a constant.

i.e. $\;$ $x + 3y = 3p$ $\;\;\; \cdots \; (3)$

Solving equations $(2)$ and $(3)$ simultaneously we have

$x = \dfrac{15 + 3p}{10}$

The asymptotes intersect at the center.

Given: Center $= \left(h, k\right) = \left(2, 1\right)$

$\implies$ $\dfrac{15 + 3p}{10} = 2$ $\implies$ $p = \dfrac{5}{3}$

Substituting the value of $p$ in equation $(3)$, the equation of the second asymptote is

$x + 3y - 5 = 0$ $\;\;\; \cdots \; (4)$

$\therefore$ $\;$ The combined equation of the asymptotes is

$\left(3x - y - 5\right) \left(x + 3y - 5\right) = 0$ $\;\;\; \cdots \; (5)$

The equation of rectangular hyperbola differs from the combined equation of asymptotes by a constant.

$\therefore$ $\;$ The equation of the rectangular hyperbola is

$\left(3x - y - 5\right) \left(x + 3y - 5\right) + \ell = 0$ $\;\;\; \cdots \; (6)$

Equation $(6)$ passes through the point $\left(1, -1\right)$.

$\therefore$ $\;$ We have

$\left(3 + 1 - 5\right) \left(1 - 3 - 5\right) + \ell = 0$ $\implies$ $\ell = -7$

$\therefore$ $\;$ The required equation of rectangular hyperbola is

$\left(3x - y - 5\right) \left(x + 3y - 5\right) - 7 = 0$

Analytical Geometry - Conics - Rectangular Hyperbola

A standard rectangular hyperbola has its vertices at $\left(5, 7\right)$ and $\left(-3, -1\right)$. Find its equation and asymptotes.

Let the required equation of the general rectangular hyperbola be

$\left(x - h\right) \left(y - k\right) = c^2$ $\;\;\; \cdots \; (1)$

The vertices of the rectangular hyperbola are $\left(5, 7\right)$ and $\left(-3, -1\right)$.

Let the center of the rectangular hyperbola be $\left(h, k\right)$.

The midpoint of the vertices of a rectangular hyperbola is its center.

$\therefore$ $\;$ We have, $\;$ $h = \dfrac{5 - 3}{2} = 1$ $\;$ and $\;$ $k = \dfrac{7 - 1}{2} = 3$

$\therefore$ $\;$ The center of the rectangular hyperbola is $\left(h, k\right) = \left(1, 3\right)$

Substituting the values of $h$ and $k$ in equation $(1)$ we have,

$\left(x -1\right) \left(y - 3\right) = c^2$ $\;\;\; \cdots \; (2)$

The vertices $\left(5, 7\right)$ and $\left(-3, -1\right)$ lie on the hyperbola.

$\therefore$ $\;$ We have from equation $(2)$,

$\left(5 - 1\right) \left(7 - 3\right) = c^2$ $\implies$ $c^2 = 16$

Substituting the value of $c^2$ in equation $(2)$, the required equation of rectangular hyperbola is

$\left(x -1\right) \left(y - 3\right) = 16$

The equation of rectangular hyperbola differs from the combined equation of asymptotes by a constant.

$\therefore$ $\;$ The combined equation of the asymptotes is $\;$ $\left(x -1\right) \left(y - 3\right) = 0$

$\therefore$ $\;$ The equations of the asymptotes are

$x -1 = 0$ $\;$ and $\;$ $y - 3 = 0$

Analytical Geometry - Conics - Rectangular Hyperbola

Find the equation of the rectangular hyperbola which which has for one of its asymptotes the line $x + 2y - 5 = 0$ and passes through the points $\left(6,0\right)$ and $\left(-3,0\right)$.

Given: One of the asymptotes of the rectangular hyperbola is

$x + 2y - 5 = 0$ $\;\;\; \cdots \; (1)$

$\therefore$ $\;$ Slope of the asymptote $= m_1 = \dfrac{-1}{2}$

The asymptotes of a rectangular hyperbola are at right angles to one another.

$\therefore$ $\;$ Slope of the second asymptote $= m_2 = \dfrac{-1}{m_1} = 2$

$\therefore$ $\;$ Let the equation of the second asymptote be

$y = 2x + k$ $\;$ where $k$ is a constant.

i.e. $\;$ $2x - y + k = 0$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ The combined equation of the asymptotes is

$\left(x + 2y - 5\right) \left(2x - y + k\right) = 0$ $\;\;\; \cdots \; (3)$

The equation of rectangular hyperbola differs from the combined equation of asymptotes by a constant.

$\therefore$ $\;$ The equation of hyperbola is

$\left(x + 2y - 5\right) \left(2x - y + k\right) + \ell = 0$ $\;\;\; \cdots \; (4)$

Equation $(4)$ passes through the points $\left(6, 0\right)$ and $\left(-3, 0\right)$.

$\therefore$ $\;$ We have

$\left(6 + 0 - 5\right) \left(12 - 0 + k\right) + \ell = 0$ $\implies$ $k + \ell = - 12$ $\;\;\; \cdots \; (5a)$

and $\left(-3 + 0 - 5\right) \left(-6 - 0 + k\right) + \ell = 0$ $\implies$ $-8k + \ell = -48$ $\;\;\; \cdots \; (5b)$

Solving equations $(5a)$ and $(5b)$ simultaneously we have,

$k = 4$ $\;$ and $\;$ $\ell = -16$

Substituting the values of $k$ and $\ell$ in equation $(4)$, the equation of the required rectangular hyperbola is

$\left(x + 2y - 5\right) \left(2x - y + 4\right) - 16 = 0$

i.e. $\left(x + 2y - 5\right) \left(2x - y + 4\right) = 16$

Analytical Geometry - Conics - Rectangular Hyperbola

Find the equation of the tangent and normal at $\left(-2, \dfrac{1}{4}\right)$ to the rectangular hyperbola $2xy - 2x - 8y - 1 = 0$

Equation of given rectangular hyperbola is

$2xy - 2x - 8y -1 = 0$

i.e. $\;$ $2xy - 2x - 8y = 1$

i.e. $\;$ $2xy - 2x - 8y + 8 = 9$

i.e. $\;$ $xy - x - 4y + 4 = \dfrac{9}{2}$

i.e. $\;$ $x \left(y - 1\right) - 4 \left(y - 1\right) = \dfrac{9}{2}$

i.e. $\;$ $\left(x - 4\right) \left(y - 1\right) = \dfrac{9}{2}$ $\;\;\; \cdots \; (1)$

The standard equation of rectangular hyperbola with center at $\left(h, k\right)$ is:

$\left(x - h\right) \left(y - k\right) = c^2$ $\;\;\; \cdots \; (2)$

Comparing equations $(1)$ and $(2)$ we have

Center $= \left(h, k\right) = \left(4, 1\right)$ $\;$ and $\;$ $c^2 = \dfrac{9}{2}$

Equation of tangent to the rectangular hyperbola at the point $\left(x_1, y_1\right)$ is

$x \left(y_1 - k\right) + y \left(x_1 - h\right) = c^2 - hk + x_1y_1$

Given: Point $\left(x_1, y_1\right) = \left(-2, \dfrac{1}{4}\right)$

$\therefore$ $\;$ The required equation of tangent is

$x \left(\dfrac{1}{4} - 1\right) + y \left(-2 - 4\right) = \dfrac{9}{2} - \left(4 \times 1\right) + \left(-2 \times \dfrac{1}{4}\right)$

i.e. $\;$ $\dfrac{-3}{4}x - 6y = 0$

i.e. $\;$ $\dfrac{-x}{4}- 2y = 0$

i.e. $\;$ $x + 8y = 0$

Equation of normal to the rectangular hyperbola at the point $\left(x_1, y_1\right)$ is

$x \left(x_1 - h\right) - y \left(y_1 - k\right) = x_1 \left(x_1 - h\right) - y_1 \left(y_1 - k\right)$

$\therefore$ $\;$ The required equation of normal is

$x \left(-2 - 4\right) - y \left(\dfrac{1}{4} - 1\right) = -2 \left(-2 - 4\right) - \dfrac{1}{4} \left(\dfrac{1}{4} - 1\right)$

i.e. $\;$ $- 6x + \dfrac{3}{4}y = \dfrac{195}{16}$

i.e. $\;$ $- 2x + \dfrac{y}{4} = \dfrac{65}{16}$

i.e. $\;$ $32x - 4y + 65 = 0$

Analytical Geometry - Conics - Rectangular Hyperbola

Find the equation of the standard rectangular hyperbola whose center is $\left(\dfrac{-1}{2}, \dfrac{-1}{2}\right)$ and which passes through the point $\left(1, \dfrac{1}{4}\right)$.

The equation of standard rectangular hyperbola with center at $\left(h, k\right)$ is:

$\left(x - h\right) \left(y - k\right) = c^2$ $\;\;\; \cdots \; (1)$

Given: Center $= \left(h, k\right) = \left(\dfrac{-1}{2}, \dfrac{-1}{2}\right)$

Putting the values of $h$ and $k$ in equation $(1)$, the required equation of rectangular hyperbola becomes

$\left(x + \dfrac{1}{2}\right) \left(y + \dfrac{1}{2}\right) = c^2$ $\;\;\; \cdots \; (2)$

Equation $(2)$ passes through the point $\left(1, \dfrac{1}{4}\right)$.

$\therefore$ $\;$ We have

$\left(1 + \dfrac{1}{2}\right) \left(\dfrac{1}{4} + \dfrac{1}{2}\right) = c^2$ $\implies$ $c^2 = \dfrac{9}{8}$

$\therefore$ $\;$ The required equation of the rectangular hyperbola [equation $(2)$] is

$\left(x + \dfrac{1}{2}\right) \left(y + \dfrac{1}{2}\right) = \dfrac{9}{8}$

Analytical Geometry - Conics - Asymptotes of Hyperbola

Find the angle between the asymptotes of the hyperbola $\;$ $24x^2 - 8y^2 = 27$

Equation of hyperbola is $\;$ $24x^2 - 8y^2 = 27$

i.e. $\;$ $\dfrac{x^2}{27 / 24} - \dfrac{y^2}{27 / 8} = 1$

i.e. $\;$ $\dfrac{x^2}{9 / 8} - \dfrac{y^2}{27 / 8} = 1$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the standard equation of hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = \dfrac{9}{8}$ $\implies$ $a = \dfrac{3}{2 \sqrt{2}}$

and $b^2 = \dfrac{27}{8}$ $\implies$ $b = \dfrac{3 \sqrt{3}}{2 \sqrt{2}}$

The angle between the asymptotes is

$\begin{aligned} 2 \alpha & = 2 \tan^{-1} \left(\dfrac{b}{a}\right) \\\\ & = 2 \tan^{-1} \left(\dfrac{3 \sqrt{3} / 2 \sqrt{2}}{3 / 2 \sqrt{2}}\right) \\\\ & = 2 \tan^{-1} \left(\sqrt{3}\right) \end{aligned}$

i.e. $\;$ $2 \alpha = \dfrac{2 \pi}{3}$

Analytical Geometry - Conics - Asymptotes of Hyperbola

Find the equation of the hyperbola if its asymptotes are parallel to $\;$ $x + 2y - 12 = 0$ $\;$ and $\;$ $x - 2y+ 8 = 0$, $\;$ $\left(2,4\right)$ is the center of the hyperbola and it passes through $\left(2,0\right)$.

Equations of the given straight lines are

$x + 2y - 12 = 0$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x - 2y + 8 = 0$ $\;\;\; \cdots \; (2)$

Slope of equation $(1)$ is $= m_1 = \dfrac{-1}{2}$

Slope of equation $(2)$ is $= m_2 = \dfrac{1}{2}$

$\because$ $\;$ the asymptotes are parallel to the given lines [equations $(1)$ and $(2)$],

$\therefore$ $\;$ the slopes of the asymptotes are $\dfrac{1}{2}$ and $\dfrac{-1}{2}$

Let the equations of the asymptotes be

$y = \dfrac{1}{2}x + p$ $\;\;\; \cdots \; (3a)$ $\;$ and $\;$ $y = -\dfrac{1}{2}x + q$ $\;\;\; \cdots \; (3b)$

where $p$ and $q$ are the Y intercepts.

Solving equations $(3a)$ and $(3b)$ simultaneously gives the point of intersection of the asymptotes.

$\therefore$ $\;$ We have from equations $(3a)$ and $(3b)$

$\dfrac{1}{2}x + p = \dfrac{-1}{2}x + q$ $\implies$ $x = q - p$

Substituting the value of $x$ in equation $(3a)$ gives

$y = \dfrac{1}{2} \left(q - p\right) + p$ $\implies$ $y = \dfrac{1}{2} \left(p + q\right)$

The asymptotes intersect at the center.

Given: Center of hyperbola $= \left(2, 4\right)$

$\therefore$ $\;$ We have $\;$ $q - p = 2$ $\;\;\; \cdots \; (4a)$

and $\;$ $\dfrac{1}{2} \left(p + q\right) = 4$ $\implies$ $p + q = 8$ $\;\;\; \cdots \; (4b)$

Solving equations $(4a)$ and $(4b)$ simultaneously we have,

$2q = 10$ $\implies$ $q = 5$

Substituting the value of $q$ in equation $(4a)$ gives

$p = q - 2 = 3$

$\therefore$ $\;$ The equations of the asymptotes are

$y = \dfrac{1}{2}x + 3$ $\;\;$ i.e. $\;$ $x - 2y + 6 = 0$

and $\;$ $y = \dfrac{-1}{2}x + 5$ $\;\;$ i.e. $\;$ $x + 2y - 10 = 0$

$\therefore$ $\;$ The combined equations of the asymptotes is

$\left(x - 2y + 6\right) \left(x + 2y - 10\right) = 0$ $\;\;\; \cdots \; (5)$

The equation of the hyperbola differs from the combined equation of the asymptotes by a constant.

$\therefore$ $\;$ The equation of the hyperbola is of the form

$\left(x - 2y + 6\right) \left(x + 2y - 10\right) + k = 0$ $\;\;\; \cdots \; (6)$

The required hyperbola passes through the point $\left(2, 0\right)$.

$\therefore$ $\;$ We have from equation $(6)$,

$\left(2 - 0 + 6\right) \left(2 + 0 - 10\right) + k = 0$

$\implies$ $k = 64$

Substituting the value of $k$ in equation $(6)$, the equation of the required hyperbola is

$\left(x - 2y + 6\right) \left(x + 2y - 10\right) + 64 = 0$

Analytical Geometry - Conics - Asymptotes of Hyperbola

Find the equation of the asymptotes to the hyperbola $8x^2 + 10xy - 3y^2 - 2x + 4y -2 = 0$

Equation of given hyperbola is $\;$ $8x^2 + 10xy - 3y^2 - 2x + 4y - 2 = 0$ $\;\;\; \cdots \; (1)$

The combined equation of the asymptotes differs from the hyperbola by a constant only.

$\therefore$ $\;$ Let the combined equation of the asymptotes be: $\;$ $8x^2 + 10xy - 3y^2 - 2x + 4y + k = 0$

Now,

$\begin{aligned} 8x^2 + 10xy - 3y^2 & = 8x^2 + 12xy - 2xy - 3y^2 \\\\ & = 4x \left(2x + 3y\right) - y \left(2x + 3y\right) \\\\ & = \left(4x - y\right) \left(2x + 3y\right) \end{aligned}$

$\therefore$ $\;$ The separate equations of the asymptotes are

$4x - y + \ell = 0$ $\;\;\; \cdots \; (2a)$ and

$2x + 3y + m = 0$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ $\left(4x - y + \ell\right) \left(2x + 3y + m\right) = 8x^2 + 10xy - 3y^2 - 2x + 4y + k$ $\;\;\; \cdots \; (3)$

Equating the coefficient of the $x$ terms we have

$4 m + 2 \ell = -2$

i.e. $\;$ $2 m + \ell = -1$ $\;\;\; \cdots \; (4a)$

Equating the coefficient of the $y$ terms we have

$- m + 3 \ell = 4$ $\;\;\; \cdots \; (4b)$

Equating the constant term we have

$\ell m = k$ $\;\;\; \cdots \; (4c)$

Solving equations $(4a)$ and $(4b)$ simultaneously we get

$\ell = 1$ $\;$ and $m = -1$

$\therefore$ $\;$ We have from equation $(4c)$, $\;$ $k = -1$

$\therefore$ $\;$ The separate equations of the asymptotes are

$4x - y + 1 = 0$ $\;$ and $\;$ $2x + 3y - 1 = 0$

The combined equation of the asymptotes is

$8x^2 + 10xy - 3y^2 - 2x + 4y - 1 = 0$

Analytical Geometry - Conics - Tangents and Normals

Find the equation of the chord of contact of tangents from the point $\left(5,3\right)$ to the hyperbola $4x^2 - 6y^2 = 24$.

Given equation of hyperbola is $\;$ $4x^2 - 6y^2 = 24$

i.e. $\;$ $\dfrac{x^2}{24 / 4} - \dfrac{y^2}{24 / 6} = 1$

i.e. $\;$ $\dfrac{x^2}{6} - \dfrac{y^2}{4} = 1$

The transverse axis of the given hyperbola is along the X axis.

Comparing with the standard equation of the hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 6$ $\;$ and $\;$ $b^2 = 4$

Equation of chord of contact of tangents from the point $\left(x_1, y_1\right)$ to the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ is: $\;$ $\dfrac{xx_1}{a^2} - \dfrac{yy_1}{b^2} = 1$

Here $\;$ $\left(x_1, y_1\right) = \left(5,3\right)$

$\therefore$ $\;$ The required equation is

$\dfrac{5x}{6} - \dfrac{3y}{4} = 1$

i.e. $\;$ $10x - 9y - 12 = 0$

Analytical Geometry - Conics - Tangents and Normals

Find the equation of the chord of contact of tangents from the point $\left(-3,1\right)$ to the parabola $y^2 = 8x$.

Equation of parabola is $\;$ $y^2 = 8x$

Comparing with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = 8 \implies a = 2$

Equation of the chord of contact of tangents from the point $\left(x_1, y_1\right)$ to the parabola $y^2 = 4ax$ is

$yy_1 = 2a \left(x + x_1\right)$

Here, $\left(x_1, y_1\right) = \left(-3,1\right)$

$\therefore$ $\;$ Required equation is

$1 \times y = 2 \times 2 \times \left(x - 3\right)$

i.e. $\;$ $4x - y - 12 = 0$

Analytical Geometry - Conics - Tangents and Normals

Show that the line $x - y + 4 = 0$ is a tangent to the ellipse $x^2 + 3y^2 = 12$. Find the coordinates of the point of contact.

Equation of given line is $\;$ $x - y + 4 = 0$

i.e. $\;$ $y = x + 4$ $\;\;\; \cdots \; (1)$

Slope of given line $= m = 1$

Intercept of given line $= c = 4$

Equation of ellipse is $\;$ $x^2 + 3y^2 = 12$

i.e. $\;$ $\dfrac{x^2}{12} + \dfrac{y^2}{12 / 3} = 1$

i.e. $\;$ $\dfrac{x^2}{12} + \dfrac{y^2}{4} = 1$ $\;\;\; \cdots \; (2)$

The major axis of the ellipse given by equation $(2)$ is along the X axis.

$\therefore$ $\;$ Comparing equation $(2)$ with the tandard equation of ellipse $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 12$ $\;$ and $\;$ $b^2 = 4$

Condition that the given line is a tangent to the ellipse is $\;$ $c^2 = a^2 m^2 + b^2$ $\;\;\; \cdots \; (3)$

Substituting the values of $c$, $a^2$, $m$ and $b^2$ in equation $(3)$ we have,

$\left(4\right)^2 = 12 \times \left(1\right)^2 + 4$

i.e. $\;$ $16 = 16$ $\;$ which is true.

$\therefore$ $\;$ The line given by equation $(1)$ is a tangent to the ellipse given by equation $(2)$.

The point of contact of the tangent with the ellipse is $\;$ $\left(\dfrac{-a^2 m}{c}, \dfrac{b^2}{c}\right)$

i.e. $\;$ $\left(\dfrac{-12 \times 1}{4}, \dfrac{4}{4}\right)$ $\;\;$ i.e. $\;$ $\left(-3, 1\right)$

Analytical Geometry - Conics - Tangents and Normals

Find the equation of the two tangents that can be drawn from the point $\left(1, 2\right)$ to the hyperbola $2x^2 - 3y^2 = 6$.

Equation of hyperbola is $\;$ $2x^2 - 3y^2 = 6$

i.e. $\;$ $\dfrac{x^2}{6 / 2} - \dfrac{y^2}{6 / 3} = 1$

i.e. $\;$ $\dfrac{x^2}{3} - \dfrac{y^2}{2} = 1$ $\;\;\; \cdots \; (1)$

The transverse axis of the hyperbola given by equation $(1)$ is along the X axis.

$\therefore$ $\;$ Comparing equation $(1)$ with the standard equation of hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 3$ $\;$ and $b^2 = 2$

Let the equation of the tangent be $\;$ $y = mx + \sqrt{a^2 m^2 - b^2}$ $\;\;\; \cdots \; (2)$

Substituting the values of $a^2$ and $b^2$ in equation $(2)$, we have,

$y = mx + \sqrt{3m^2 - 2}$ $\;\;\; \cdots \; (3)$

Equation $(3)$ passes through the point $\left(1, 2\right)$.

$\therefore$ $\;$ We have from equation $(3)$,

$2 = m + \sqrt{3m^2 - 2}$

i.e. $\;$ $2 - m = \sqrt{3m^2 - 2}$ $\;\;\; \cdots \; (4)$

Squaring both sides of equation $(4)$ we get,

$4 + m^2 - 4m = 3m^2 - 2$

i.e. $\;$ $2m^2 + 4m - 6 = 0$

i.e. $\;$ $m^2 + 2m - 3 = 0$

i.e. $\;$ $\left(m + 3\right) \left(m - 1\right) = 0$

i.e. $\;$ $m = -3$ $\;$ or $\;$ $m = 1$

Substituting $m = -3$ in equation $(3)$ gives

$y = -3x + \sqrt{3 \times \left(-3\right)^2 - 2}$

i.e. $\;$ $3x + y - 5 = 0$ $\;\;\; \cdots \; (5a)$

Substituting $m = 1$ in equation $(3)$ gives

$y = x + \sqrt{3 \times \left(1\right)^2 - 2}$

i.e. $\;$ $x - y + 1 = 0$ $\;\;\; \cdots \; (5b)$

Equations $(5a)$ and $(5b)$ are the required equations of tangents.

Analytical Geometry - Conics - Tangents and Normals

Find the equation of the two tangents that can be drawn from the point $\left(2,-3\right)$ to the parabola $y^2 = 4x$.

Equation of parabola is $\;$ $y^2 = 4x$ $\;\;\; \cdots \; (1)$

Comparing with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives $\;$ $a = 1$

Let the equation of tangent be $\;$ $y = mx + \dfrac{a}{m}$ $\;\;\; \cdots \; (2)$

Substituting the value of $a$ in equation $(2)$ gives

$y = mx + \dfrac{1}{m}$ $\;\;\; \cdots \; (3)$

Equation $(3)$ passes through the point $\left(2, -3\right)$.

$\therefore$ $\;$ We have $\;$ $-3 = 2m + \dfrac{1}{m}$

i.e. $\;$ $2m^2 + 3m + 1 = 0$

i.e. $\;$ $\left(2m + 1\right) \left(m + 1\right) = 0$

i.e. $\;$ $m = -1$ $\;$ or $\;$ $m = \dfrac{-1}{2}$

When $m = -1$, we have from equation $(3)$, $\;$ $y = -x -1$

i.e. $\;$ $x + y + 1 = 0$ $\;\;\; \cdots \; (4a)$

When $m = \dfrac{-1}{2}$, we have from equation $(3)$, $\;$ $y = \dfrac{-x}{2} - \dfrac{1}{1 / 2}$

i.e. $\;$ $y = \dfrac{-x}{2} - 2$

i.e. $\;$ $x + 2y + 4 = 0$ $\;\;\; \cdots \; (4b)$

Equations $(4a)$ and $(4b)$ are the required equations of tangents.

Analytical Geometry - Conics - Tangents and Normals

Find the equation of tangents to the hyperbola $4x^2 - y^2 = 64$, which are parallel to $10x-3y+9=0$.

Equation of hyperbola is $\;$ $4x^2 - y^2 = 64$

i.e. $\;$ $\dfrac{x^2}{64 / 4} - \dfrac{y^2}{64} = 1$

i.e. $\;$ $\dfrac{x^2}{16} - \dfrac{y^2}{64} = 1$ $\;\;\; \cdots \; (1)$

The transverse axis of the hyperbola given by equation $(1)$ is along the X axis.

Comparing equation $(1)$ with the standard equation of hyperbola $\;$ $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 16 \implies a = 4$ $\;$ and $\;$ $b^2 = 64 \implies b = 8$

Equation of given line is $\;$ $10x - 3y + 9 = 0$

i.e. $\;$ $y = \dfrac{10}{3} x + 3$

Slope of given line $= m = \dfrac{10}{3}$

$\because$ $\;$ the required tangents are parallel to the given line,

$\therefore$ $\;$ slope of required tangents $= m = \dfrac{10}{3}$

Equations of tangents (with slope $m$) to the hyperbola are $\;$ $y = mx \pm \sqrt{a^2 m^2 - b^2}$

$\therefore$ $\;$ Required equations of tangents to the hyperbola are

$y = \dfrac{10}{3} x \pm \sqrt{16 \times \dfrac{100}{9} - 64}$

i.e. $\;$ $y = \dfrac{10}{3} x \pm \sqrt{\dfrac{1024}{9}}$

i.e. $\;$ $y = \dfrac{10}{3} x \pm \dfrac{32}{3}$

i.e. $\;$ $10 x - 3y \pm 32 = 0$

Analytical Geometry - Conics - Tangents and Normals

Find the equation of tangents to the ellipse $\dfrac{x^2}{20}+ \dfrac{y^2}{5} = 1$, which are perpendicular to $x+y+2=0$.

Equation of ellipse is $\;$ $\dfrac{x^2}{20} + \dfrac{y^2}{5} = 1$ $\;\;\; \cdots \; (1)$

The major axis of the ellipse given by equation $(1)$ is along the X axis.

Comparing equation $(1)$ with the standard equation of ellipse $\;$ $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ $\;$ gives

$a^2 = 20$ $\;$ and $\;$ $b^2 = 5$

Equation of given line is $\;$ $x + y + 2 = 0$

i.e. $\;$ $y = -x - 2$

Slope of given line $= m_1 = -1$

$\because$ $\;$ the required tangents are perpendicular to the given line,

slope of tangent $= m = \dfrac{-1}{m_1} = 1$

Equations of tangents (with slope $m$) to the ellipse are $\;$ $y = mx \pm \sqrt{a^2 m^2 + b^2}$

$\therefore$ $\;$ Required equations of tangents to the ellipse are

$y = 1 \times x \pm \sqrt{20 \times 1^2 + 5}$

i.e. $\;$ $y = x \pm 5$

Analytical Geometry - Conics - Tangents and Normals

Find the equation of tangent to the parabola $y^2 = 6x$, parallel to $3x - 2y + 5 = 0$.

Equation of parabola is $\;$ $y^2 = 6x$

Comparing with the standard equation of parabola $\;$ $y^2 = 4ax$ $\;$ gives

$4a = 6 \implies a = \dfrac{3}{2}$

Equation of given line is $\;$ $3x - 2y + 5 = 0$

i.e. $\;$ $y = \dfrac{3}{2}x + \dfrac{5}{2}$

$\therefore$ $\;$ Slope of the given line $= m = \dfrac{3}{2}$

$\because$ $\;$ the tangent to the parabola is is parallel to the given line, slope of tangent to the parabola $= m = \dfrac{3}{2}$

The equation of tangent (with slope m) to the parabola $y^2 = 4ax$ is $\;$ $y = mx + \dfrac{a}{m}$

$\therefore$ $\;$ the required equation of tangent is

$y = \dfrac{3}{2} x + \dfrac{3 / 2}{3 / 2}$

i.e. $\;$ $y = \dfrac{3}{2}x + 1$

i.e. $\;$ $3x - 2y + 2 = 0$